Truth of $x^2-2=0$

From Algrebra by Gelfand, it says that "When we claim that we have solved the equation $x^2-2=0$ and the answer is $x=\sqrt{2}$ or $x=-\sqrt{2}$, we are in fact cheating. To tell the truth, we have not solved this equation but confessed our inability to solve it; $\sqrt{2}$ means nothing except "the positive solution of the equation $x^2-2=0$

Can someone please explain what he really means? Does he mean every irrational number is meaningless? Thanks in advance.

Solution 1:

He is saying the following.

1. $\sqrt{2}$ is defined as "the number such that its square is two."
2. The statement "the solution $x$ to $x^2-2 = 0$ is $\sqrt{2}$" is therefore saying "the solution $x$ to $x^2-2=0$ is the number such that its square is two."

As you can see, this is a rather circular statement. This doesn't mean that irrational numbers are meaningless (indeed, $\sqrt{2}$ does exist -- see the comments and @EthanBolker's answer for more on this), but rather saying that this method of definition limits us to statements like "the number such that its square is two" or "half of the number such that its square is eight."

Solution 2:

To elaborate on @fractal1729 's lovely correct answer:

Having defined $\sqrt{2}$ as "the number such that it's square is $2$" it's reasonable to ask whether there is such a number.

The Greeks knew that the answer is "no" if "number" means "rational number". So in order to claim the existence of $\sqrt{2}$ you must enlarge the system of rational numbers. The Greeks essentially sidestepped that question by doing geometry rather than arithmetic, using points on a line instead of numbers. Clearly there's a point on the "number line" with the right length since you can draw the diagonal of a unit square.

In later centuries mathematicians found more formal algebraic ways to enlarge the system of rational numbers to include what we now call all the "real numbers". The new ones are the irrationals. There are several ways to do this. One of the most common is to make precise the notion of an infinite decimal, along with rules for arithmetic with them. Others come with the names "Cauchy sequences" or "Dedekind cuts".

Solution 3:

Using definite descriptor notation, we can define:

$$\sqrt{x} = (\iota y \in \mathbb{R})(y \geq 0 \wedge y^2=x)$$

In words: $\sqrt{x}$ is the unique $y \geq 0$ such that $y^2=x$.

From this, you can show that for all $y$ and all $x \geq 0$, we have: $$y^2 = x \iff y \in \pm\sqrt{x}.$$

But in some sense, this is a purely logical construct, at least insofar as we haven't explained how to approximate $\sqrt{x}$ to arbitrary precision. There's ways of doing this, but we haven't given one.

Solution 4:

I think the author is simply wrong. There is another meaningful way of defining $\sqrt{2}$. Consider the following sequence: $$x_{n+1} := \frac{1}{2}\left(x_n + \frac{2}{x_n}\right)$$ with $x_0=1$.

Then $\lim_{n\to\infty} x_n=\sqrt{2}$, even though all $x_n\in\mathbb{Q}$ for arbitrary $n\in\mathbb{N}$. Now instead of defining $\sqrt{2}$ as the solution of $x^2-2=0$ we could have also defined it as the limit on $x_n$.

Proof for the sequence can be found here (German only).